Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $k = \dfrac{p^2 - 5p - 50}{10p - 90} \times \dfrac{-2p + 18}{3p^2 + 15p} $
Answer: First factor the quadratic. $k = \dfrac{(p + 5)(p - 10)}{10p - 90} \times \dfrac{-2p + 18}{3p^2 + 15p} $ Then factor out any other terms. $k = \dfrac{(p + 5)(p - 10)}{10(p - 9)} \times \dfrac{-2(p - 9)}{3p(p + 5)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ (p + 5)(p - 10) \times -2(p - 9) } { 10(p - 9) \times 3p(p + 5) } $ $k = \dfrac{ -2(p + 5)(p - 10)(p - 9)}{ 30p(p - 9)(p + 5)} $ Notice that $(p - 9)$ and $(p + 5)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ -2\cancel{(p + 5)}(p - 10)(p - 9)}{ 30p(p - 9)\cancel{(p + 5)}} $ We are dividing by $p + 5$ , so $p + 5 \neq 0$ Therefore, $p \neq -5$ $k = \dfrac{ -2\cancel{(p + 5)}(p - 10)\cancel{(p - 9)}}{ 30p\cancel{(p - 9)}\cancel{(p + 5)}} $ We are dividing by $p - 9$ , so $p - 9 \neq 0$ Therefore, $p \neq 9$ $k = \dfrac{-2(p - 10)}{30p} $ $k = \dfrac{-(p - 10)}{15p} ; \space p \neq -5 ; \space p \neq 9 $